A capacitor has a capacitance of 7.28uF. What amount of excess charge must be placed on each of its plates to make the potential difference between the plates equal to 25.0V?
charge = capacitance * voltage
coulombs = Farads * volts
To determine the amount of excess charge on each plate of a capacitor, we need to use the formula:
Q = C × V
where:
Q is the amount of charge (in coulombs)
C is the capacitance (in farads)
V is the potential difference (in volts)
In this case, we are given:
C = 7.28 μF (microfarads, or 7.28 × 10^(-6) F)
V = 25.0 V
To find the amount of charge, we substitute the given values into the formula:
Q = (7.28 × 10^(-6) F) × (25.0 V)
Calculating Q:
Q = 1.82 × 10^(-4) C
Hence, the amount of excess charge on each plate of the capacitor is 1.82 × 10^(-4) C (coulombs).